Question

If the equation of projection is `y=ax-bx^2`, find initial velocity(`u`) and angle of projection(`theta`)?

Answer 1

Given that the equation of projection is `y=ax-bx^2`, we are to find initial velocity(`u`) and angle of projection(`theta`)

Horizontal component of velocity of projection is `ucostheta`

Vertical component of velocity of projection is `usintheta`

Let at the t th sec after its projection from origin its position be represented by the coordinates `(x,y)`

So `x=ucosthetaxxt`

`=>t=x/(ucostheta).....[1]`

Again

`y=usinthetaxxt-1/2gxxt^2.....[2]`

Combining [1] and[2] we get

`y=usinthetaxxx/(ucostheta)-1/2gxx x^2/(u^2cos^2theta)`

`=>y=xtantheta- (gx^2)/(2u^2cos^2theta)....[3]`

Comparing equation [3] with the given equation we get

`a=tantheta`

`=>theta=tan^-1a`

and

`b=g/(2u^2cos^2theta)`

`=>u^2=g/(2b)sec^2theta`

`=>u^2=g/(2b)(1+tan^2theta)`

`=>u^2=g/(2b)(1+a^2)`

`=>u=sqrt(g/(2b)(1+a^2))`

Answer 2

See below.

Explanation:

This is a typical kinematic problem. Assuming that the movement begins at the referential origin we have

`{(x = v_x t),(y=v_y t-1/2 g t^2):}`

where

`t` is the elapsed time
`v_x` the `x`-axis velocity component
`v_y` the `y`-axis velocity component
`g` the gravity acceleration.

Now the orbit equation is

`y-ax+bx^2=0` substituting the parametric equations we have

`1/2(g t + 2 a v_x - 2 b t v_x^2 - 2 v_y)t=0`

This condition must be true for all `t` so

`(g t + 2 a v_x - 2 b t v_x^2 - 2 v_y)=0, forall t`

Now grouping coefficients

`(g - 2 b v_x^2)t+2(a v_x - v_y)=0` so

`{(g - 2 b v_x^2=0),(a v_x - v_y=0):}`

Now solving for `v_x, v_y`

`{(v_x=sqrt(g/(2b))),(v_y=a sqrt(g/(2b))):}`

Now `v=sqrt(v_x^2+v_y^2) = sqrt((g/2)((a^2+1)/b))`

`theta = arctan(v_y/v_x) =arctan(a)`